JEE Physics Centre of Mass MCQs Set 11

Answer: (b) \( \left(\frac{11}{3},\frac{10}{3}\right) \)

Explanation:
1. \( x = \frac{5(3)+10(4)}{15} = \frac{11}{3} \), \( y = \frac{5(4)+10(3)}{15} = \frac{10}{3} \).

Teacher's Note:
1. Weight positions by mass.
2. Divide by total mass 15.

Answer: (d) \( 2a\hat i+\frac{4a}{3}\hat j \)

Explanation:
1. Masses are proportional to 2σ, 3σ, σ with centres at (a, a), (3a, a), (a, 3a).
2. \( x = \frac{2(a)+3(3a)+1(a)}{6} = 2a \), \( y = \frac{2(a)+3(a)+1(3a)}{6} = \frac{4a}{3} \).

Teacher's Note:
1. Equal areas, so masses go as density.
2. Use each plate's own centre.

Answer: (a) \( \frac53 \) m, in the direction in which the 50 kg man moves (to the right if the 50 kg man starts at the left)

Explanation:
1. The centre of mass of the system stays fixed. Relative to the plank the men's centre of mass shifts by \( \frac{1000-500}{150} = \frac{10}{3} \) m towards the left.
2. \( 150\,d+150\left(d-\frac{10}{3}\right)=0 \Rightarrow d = \frac53 \) m to the right.

Teacher's Note:
1. No external horizontal force.
2. The plank moves opposite to the men's centre of mass shift.

Answer: (c) \( \frac52 \) m/s² downward

Explanation:
1. Atwood machine: \( a = \frac{15-5}{20}g = 5 \) m/s².
2. \( a_{cm} = \frac{15(5\downarrow)+5(5\uparrow)}{20} = 2.5 \) m/s² downward.

Teacher's Note:
1. Masses accelerate in opposite directions.
2. Take vector sum weighted by mass.

Answer: (c) 5 cm

Explanation:
1. For a solid cone the centre of mass is at \( \frac h4 \) from the base: \( \frac{20}{4} = 5 \) cm.

Teacher's Note:
1. Solid cone: h/4 from base.
2. Hollow cone: h/3.

Answer: (a) \( \frac{3L}{4} \)

Explanation:
1. \( x_{cm} = \frac{\int_0^L x\,kx^2dx}{\int_0^Lkx^2dx} = \frac{L^4/4}{L^3/3} = \frac{3L}{4} \).

Teacher's Note:
1. Weight x by dm = λdx.
2. Denser end is farther.

Answer: (d) zero

Explanation:
1. No external force acts and the initial momentum is zero, so the centre of mass stays at rest.

Teacher's Note:
1. Internal forces do not move the centre of mass.
2. Total momentum stays zero.

Answer: (b) \( 15\sqrt2 \) m/s

Explanation:
1. Masses 0.25, 0.25 and 0.5 kg. Momentum of the two equal parts: \( 0.25\times30\times\sqrt2 = 7.5\sqrt2 \).
2. The bigger part has equal and opposite momentum: \( 0.5v = 7.5\sqrt2 \Rightarrow v = 15\sqrt2 \) m/s.

Teacher's Note:
1. Initial momentum zero.
2. Combine perpendicular momenta by Pythagoras.

Answer: (b) \( \frac{-a^3b}{R^3-a^3} \)

Explanation:
1. Masses are proportional to volumes: removed \( a^3 \), remaining \( R^3-a^3 \).
2. \( x = \frac{0-a^3b}{R^3-a^3} \).

Teacher's Note:
1. Hole is a negative mass.
2. Mass ∝ r³ for spheres.

Answer: (c) \( \frac R6 \)

Explanation:
1. Hole radius R/2 with its centre at R/2 from O. Areas: removed \( \frac{\pi R^2}{4} \), remaining \( \frac{3\pi R^2}{4} \).
2. \( x = \frac{-(R/2)(1/4)}{3/4} = -\frac R6 \), so the distance is R/6.

Teacher's Note:
1. Hole area is a quarter of the disc.
2. Treat the hole as negative mass.

Answer: (a) \( \left(\frac{4R}{3\pi},\frac{4R}{3\pi}\right) \)

Explanation:
1. A semicircular disc has its centre of mass at \( \frac{4R}{3\pi} \) from the diameter. By symmetry a quarter disc has \( x_{cm} = y_{cm} = \frac{4R}{3\pi} \).

Teacher's Note:
1. Standard result for lamina.
2. Symmetry about the line y = x.

Answer: (d) 7.5 R

Explanation:
1. Their surfaces touch when the centres are 3R apart, so the centres approach by \( 12R-3R = 9R \).
2. The centre of mass stays fixed, so distances moved are inversely proportional to masses: smaller body moves \( \frac{5}{6}\times9R = 7.5R \).

Teacher's Note:
1. No external force.
2. M x₁ = 5M x₂.

Answer: (b) 112 m

Explanation:
1. The centre of mass still lands at 100 m: \( \frac{1(40)+5x}{6} = 100 \).
2. \( x = 112 \) m.

Teacher's Note:
1. Fragments' centre of mass follows the original path.
2. Weight positions by mass.

Answer: (d) \( \sqrt{20} \) m/s²

Explanation:
1. \( a = \frac{100-0.2\times50}{15} = 6 \) m/s² for each block.
2. \( \vec a_{cm} = \frac{5(6)\hat i+10(6)\hat j}{15}=2\hat i+4\hat j \), magnitude \( \sqrt{20} \) m/s².

Teacher's Note:
1. The blocks accelerate in perpendicular directions.
2. Find a from Newton's laws first.

Answer: (a) 0.8

Explanation:
1. \( e = \sqrt{\frac{h_2}{h_1}} = \sqrt{\frac{16}{25}} = 0.8 \).

Teacher's Note:
1. e = ratio of separation speed to approach speed.
2. Speeds go as √h.